Equilibrium constant, Kc?

Question

Consider the following reaction:

Fe^3+(aq) + SCN^-(aq) <---> FeSCN^2+(aq)

A solution is made containing an initial [Fe^3+] of 1.2×10−3 M and an initial [SCN^-] of 7.9×10−4 M. At equilibrium, [FeSCN^2+] = 1.8×10−4 M.

Calculate the value of the equilibrium constant (Kc).

Any help would be appreciated. Thanks.

Answer 1

Kc=[Fe(SCN)^2+]/[Fe^3+][SCN-]

=[1.8x10^-4]/[1.2x10^-3-1.8x10^-4] [7.9x10^-4-1.8x10^-4]

=289

Answer 2

I used a reaction table to solve this one.

.................. Fe3+(aq).... +..... SCN-(aq)... --&gt;..... FeSCN2+(aq)

Initial........ 1.2*10^-3M ........ 7.9*10^-4M ................. 0

Reacting....... -x ......................... -x ........................... +x

Equilibrium 1.2*10^-3 -x ..... 7.9*10^-4 -x .............. 1.8*10^-4M

They give you x so its all good.

x = 1.8*10^-4

[Fe3+] = 1.2*10^-3 - x

[Fe3+] = 1.2*10^-3 - 1.8*10^-4

[Fe3+] = 1.02*10^-3M

[SCN-] = 7.9*10^-4 - x

[SCN-] = 7.9*10^-4 - 1.8*10^-4

[SCN-] = 6.1*10^-4M

Now use the equation to write the equilibrium expression and plug in the values.

Kc = [FeSCN2+]/([Fe3+][SCN-])

Kc = (1.8*10^-4) / ((1.02*10^-3)(6.1*10^-4))

Kc = 289